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1:
a: \(=\dfrac{-4}{7}+\dfrac{4}{7}+\dfrac{3}{7}-\dfrac{23}{34}-\dfrac{4}{5}=\dfrac{3}{7}-\dfrac{23}{34}-\dfrac{4}{5}=-\dfrac{1247}{1190}\)
b:
Sửa đề: \(\dfrac{-5}{13}+\dfrac{4}{19}+\dfrac{-8}{13}+\dfrac{15}{19}+\dfrac{45}{6}\)
\(=\dfrac{-5}{13}-\dfrac{8}{13}+\dfrac{4}{19}+\dfrac{15}{19}+\dfrac{45}{6}=\dfrac{9}{2}\)
\(15-2.x=25\)
\(2.x=15-25=-10\)
\(x=\left(-10\right)\div2=-5\)
\(31-\left(17+x\right)=18\)
\(\left(17+x\right)=31-18=13\)
\(x\) \(=13-17=-4\)
15−2.x=2515−2.x=25
2.x=15−25=−102.x=15−25=−10
x=(−10)÷2=−5x=(−10)÷2=−5
31−(17+x)=1831−(17+x)=18
(17+x)=31−18=13(17+x)=31−18=13
xx =13−17=−4
a) \(0,6+\dfrac{2}{3}=\dfrac{6}{10}+\dfrac{2}{3}=\dfrac{3}{5}+\dfrac{2}{3}=\dfrac{9}{15}+\dfrac{10}{15}=\dfrac{19}{15}\)
b) \(-\dfrac{5}{12}+0,75=-\dfrac{5}{12}+\dfrac{75}{100}=-\dfrac{5}{12}+\dfrac{3}{4}=-\dfrac{5}{12}+\dfrac{9}{12}=\dfrac{4}{12}=\dfrac{1}{3}\)
c) \(\dfrac{1}{3}-\left(-0,4\right)=\dfrac{1}{3}+\dfrac{4}{10}=\dfrac{1}{3}+\dfrac{2}{5}=\dfrac{5}{15}+\dfrac{6}{15}=\dfrac{11}{15}\)
d) \(1\dfrac{3}{5}+\dfrac{5}{6}=\dfrac{8}{5}+\dfrac{5}{6}=\dfrac{48}{40}+\dfrac{25}{30}=\dfrac{73}{30}\)
A, \(2\frac{2}{5}\left(\frac{1}{2}x-0,75\right)=\frac{3}{10}\)
\(=>\frac{2.5+2}{5}\left(\frac{1}{2}x-\frac{3}{4}\right)=\frac{3}{10}\)
\(=>\frac{1}{2}x-\frac{3}{4}=\frac{3}{10}:\frac{12}{5}=\frac{1}{8}\)
\(=>x=\left(\frac{1}{8}+\frac{3}{4}\right):\frac{1}{2}\)
\(=>x=\frac{7}{4}\)
B, \(\frac{3}{5}-|x-\frac{1}{2}|=25\%\)
\(=>|x-\frac{1}{2}|=\frac{3}{5}-\frac{1}{4}\)
\(=>|x-\frac{1}{2}|=\frac{7}{20}\)
\(=>x-\frac{1}{2}=\frac{7}{20};-\frac{7}{20}\)
TH1: \(x-\frac{1}{2}=\frac{7}{20}=>x=\frac{17}{20}\)
TH2: \(x-\frac{1}{2}=-\frac{7}{20}=>x=\frac{3}{20}\)
(-5/24+0,75+7/12).(-2 1/4)
=(-5/24+3/4+7/12).(-9/4)
=9/8.(-9/4)
=-81/32
a, 4= 22 ; 10= 2 x 5
=> BCNN(4;10)= 22 x 5=20
b, 14=2 x 7 ;
=> BCNN(13;14)= 2 x 7 x 13= 182
c, 14=7 x 2; 21=7 x 3
=> BCNN(7;14;21)= 7 x 2 x 3 = 42
d, 15= 3 x 5 ; 18 = 2 x 32 ; 20=22 x 5
=> BCNN(15;18;20)= 32 x 22 x 5 = 180
VD9
a, 8=23 ; 12 = 22 x 3
=> BCNN(8;12)= 23 x 3= 24
b, 30 = 2 x 3 x 5; 4=22
=> BCNN(30;4)= 22 x 3 x 5 = 60
c, 20= 22 x 5
=> BCNN(2;5;20)= 22 x 5=20
d, 6=2 x 3; 14= 2x 7; 120 = 23 x 3 x 5
=> BCNN(6;14;120)= 23 x 3 x 5 x 7=840
e, 30=2 x 3 x 5 ; 6=2 x 3
=> BCNN(30;6)= 2 x 3 x 5= 30
f, 15=3 x 5; 18= 2 x 32
=> BCNN(15;18)= 2 x 32 x 5 = 90
g, 10 = 2 x 5; 24 = 23 x 3; 32= 25
=> BCNN(10;24;32)= 25 x 3 x 5 = 480
`0,75+(-1/3)-5/18`
`=3/4-1/3-5/18`
`=27/36-12/36-10/36`
`=5/36`
\(=\dfrac{3}{4}-\dfrac{1}{3}-\dfrac{5}{18}=\dfrac{27-12-10}{36}=\dfrac{5}{36}\)
`@V.Tr.V`